3.317 \(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=196 \[ \frac {a^2 (18 A+20 B+25 C) \tan (c+d x)}{15 d}+\frac {a^2 (6 A+7 B+8 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (18 A+25 B+20 C) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac {a^2 (6 A+7 B+8 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {(2 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{20 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

[Out]

1/8*a^2*(6*A+7*B+8*C)*arctanh(sin(d*x+c))/d+1/15*a^2*(18*A+20*B+25*C)*tan(d*x+c)/d+1/8*a^2*(6*A+7*B+8*C)*sec(d
*x+c)*tan(d*x+c)/d+1/60*a^2*(18*A+25*B+20*C)*sec(d*x+c)^2*tan(d*x+c)/d+1/20*(2*A+5*B)*(a^2+a^2*cos(d*x+c))*sec
(d*x+c)^3*tan(d*x+c)/d+1/5*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d

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Rubi [A]  time = 0.51, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {3043, 2975, 2968, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {a^2 (18 A+20 B+25 C) \tan (c+d x)}{15 d}+\frac {a^2 (6 A+7 B+8 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (18 A+25 B+20 C) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac {a^2 (6 A+7 B+8 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {(2 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{20 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(a^2*(6*A + 7*B + 8*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(18*A + 20*B + 25*C)*Tan[c + d*x])/(15*d) + (a^2*(6
*A + 7*B + 8*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^2*(18*A + 25*B + 20*C)*Sec[c + d*x]^2*Tan[c + d*x])/(60*
d) + ((2*A + 5*B)*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (A*(a + a*Cos[c + d*x])^2*Sec
[c + d*x]^4*Tan[c + d*x])/(5*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^2 (a (2 A+5 B)+a (2 A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx}{5 a}\\ &=\frac {(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x)) \left (a^2 (18 A+25 B+20 C)+2 a^2 (6 A+5 B+10 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac {(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (a^3 (18 A+25 B+20 C)+\left (2 a^3 (6 A+5 B+10 C)+a^3 (18 A+25 B+20 C)\right ) \cos (c+d x)+2 a^3 (6 A+5 B+10 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx}{20 a}\\ &=\frac {a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\int \left (15 a^3 (6 A+7 B+8 C)+4 a^3 (18 A+20 B+25 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{60 a}\\ &=\frac {a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^2 (6 A+7 B+8 C)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{15} \left (a^2 (18 A+20 B+25 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^2 (6 A+7 B+8 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^2 (6 A+7 B+8 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (18 A+20 B+25 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {a^2 (6 A+7 B+8 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 (18 A+20 B+25 C) \tan (c+d x)}{15 d}+\frac {a^2 (6 A+7 B+8 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (18 A+25 B+20 C) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(2 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [B]  time = 5.35, size = 502, normalized size = 2.56 \[ \frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\frac {16 (18 A+20 B+25 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 (39 A+20 (2 B+C)) \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {16 (18 A+20 B+25 C) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+\frac {129 A+145 B+140 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {-129 A-5 (29 B+28 C)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 (39 A+20 (2 B+C)) \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}-30 (6 A+7 B+8 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 (6 A+7 B+8 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 (12 A+5 B)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {3 (12 A+5 B)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {12 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}+\frac {12 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5}\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-30*(6*A + 7*B + 8*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
30*(6*A + 7*B + 8*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*(12*A + 5*B))/(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^4 + (129*A + 145*B + 140*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (12*A*Sin[(c + d*x)/2])/(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2])^5 + (2*(39*A + 20*(2*B + C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*
x)/2])^3 + (16*(18*A + 20*B + 25*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (12*A*Sin[(c + d
*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5 - (3*(12*A + 5*B))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (
2*(39*A + 20*(2*B + C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (-129*A - 5*(29*B + 28*C))
/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (16*(18*A + 20*B + 25*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])))/(960*d)

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fricas [A]  time = 0.46, size = 180, normalized size = 0.92 \[ \frac {15 \, {\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (18 \, A + 20 \, B + 25 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (6 \, A + 7 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (9 \, A + 10 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 24 \, A a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(15*(6*A + 7*B + 8*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(6*A + 7*B + 8*C)*a^2*cos(d*x + c)^5
*log(-sin(d*x + c) + 1) + 2*(8*(18*A + 20*B + 25*C)*a^2*cos(d*x + c)^4 + 15*(6*A + 7*B + 8*C)*a^2*cos(d*x + c)
^3 + 8*(9*A + 10*B + 5*C)*a^2*cos(d*x + c)^2 + 30*(2*A + B)*a^2*cos(d*x + c) + 24*A*a^2)*sin(d*x + c))/(d*cos(
d*x + c)^5)

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giac [A]  time = 1.38, size = 341, normalized size = 1.74 \[ \frac {15 \, {\left (6 \, A a^{2} + 7 \, B a^{2} + 8 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (6 \, A a^{2} + 7 \, B a^{2} + 8 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (90 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 420 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 490 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 560 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 864 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1120 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 540 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 790 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1040 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 390 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(6*A*a^2 + 7*B*a^2 + 8*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(6*A*a^2 + 7*B*a^2 + 8*C*a^2)*
log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(90*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 105*B*a^2*tan(1/2*d*x + 1/2*c)^9 + 1
20*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 420*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 490*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 560*C*
a^2*tan(1/2*d*x + 1/2*c)^7 + 864*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 1120*C*a^2*
tan(1/2*d*x + 1/2*c)^5 - 540*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 790*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 1040*C*a^2*tan(
1/2*d*x + 1/2*c)^3 + 390*A*a^2*tan(1/2*d*x + 1/2*c) + 375*B*a^2*tan(1/2*d*x + 1/2*c) + 360*C*a^2*tan(1/2*d*x +
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 0.51, size = 315, normalized size = 1.61 \[ \frac {6 a^{2} A \tan \left (d x +c \right )}{5 d}+\frac {3 a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {7 a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {7 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {5 a^{2} C \tan \left (d x +c \right )}{3 d}+\frac {a^{2} A \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {4 a^{2} B \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

6/5*a^2*A*tan(d*x+c)/d+3/5*a^2*A*sec(d*x+c)^2*tan(d*x+c)/d+7/8*a^2*B*sec(d*x+c)*tan(d*x+c)/d+7/8/d*B*a^2*ln(se
c(d*x+c)+tan(d*x+c))+5/3/d*a^2*C*tan(d*x+c)+1/2*a^2*A*sec(d*x+c)^3*tan(d*x+c)/d+3/4*a^2*A*sec(d*x+c)*tan(d*x+c
)/d+3/4/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+4/3*a^2*B*tan(d*x+c)/d+2/3*a^2*B*sec(d*x+c)^2*tan(d*x+c)/d+1/d*a^2*C
*sec(d*x+c)*tan(d*x+c)+1/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+1/5/d*a^2*A*tan(d*x+c)*sec(d*x+c)^4+1/4*a^2*B*sec(d
*x+c)^3*tan(d*x+c)/d+1/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.77, size = 360, normalized size = 1.84 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 30 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, C a^{2} \tan \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*A*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 30*A*a^2*
(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*l
og(sin(d*x + c) - 1)) - 15*B*a^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1
) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(s
in(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 120*C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) + 240*C*a^2*tan(d*x + c))/d

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mupad [B]  time = 2.20, size = 286, normalized size = 1.46 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,A}{4}+\frac {7\,B}{8}+C\right )}{d}-\frac {\left (\frac {3\,A\,a^2}{2}+\frac {7\,B\,a^2}{4}+2\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-7\,A\,a^2-\frac {49\,B\,a^2}{6}-\frac {28\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,A\,a^2}{5}+\frac {40\,B\,a^2}{3}+\frac {56\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-9\,A\,a^2-\frac {79\,B\,a^2}{6}-\frac {52\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a^2}{2}+\frac {25\,B\,a^2}{4}+6\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (6\,A+7\,B+8\,C\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^6,x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2) + 1)*((3*A)/4 + (7*B)/8 + C))/d - (tan(c/2 + (d*x)/2)^9*((3*A*a^2)/2 + (7*B*a^2)/4
+ 2*C*a^2) - tan(c/2 + (d*x)/2)^7*(7*A*a^2 + (49*B*a^2)/6 + (28*C*a^2)/3) - tan(c/2 + (d*x)/2)^3*(9*A*a^2 + (
79*B*a^2)/6 + (52*C*a^2)/3) + tan(c/2 + (d*x)/2)^5*((72*A*a^2)/5 + (40*B*a^2)/3 + (56*C*a^2)/3) + tan(c/2 + (d
*x)/2)*((13*A*a^2)/2 + (25*B*a^2)/4 + 6*C*a^2))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(
c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1)) - (a^2*log(tan(c/2 + (d*x)/2) - 1)*(6*
A + 7*B + 8*C))/(8*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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